# How do you find all the zeros of f(x)=x^3-3x^2+x-3?

Nov 2, 2016

$f \left(x\right)$ has zeros $x = 3$ and $x = \pm i$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We use this later with $a = x$ and $b = i$

$f \left(x\right) = {x}^{3} - 3 {x}^{2} + x - 3$

Note that the ratio of the first and second terms is the same as that of the third and fourth terms, so this cubic will factor by grouping:

${x}^{3} - 3 {x}^{2} + x - 3 = \left({x}^{3} - 3 {x}^{2}\right) + \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + x - 3} = {x}^{2} \left(x - 3\right) + 1 \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + x - 3} = \left({x}^{2} + 1\right) \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + x - 3} = \left({x}^{2} - {i}^{2}\right) \left(x - 3\right)$

$\textcolor{w h i t e}{{x}^{3} - 3 {x}^{2} + x - 3} = \left(x - i\right) \left(x + i\right) \left(x - 3\right)$

Hence zeros:

$x = \pm i$

$x = 3$